Dimensional Analysis: Combining 3 Variables

AI Thread Summary
The discussion centers on the challenge of combining three variables—R, μ, and dp/dx—into a dimensionless product. The user confirms that their dimensional analysis approach is correct, noting the dimensions of each variable and setting up a system of equations to demonstrate that these variables cannot yield a dimensionless parameter. They conclude that the only solution requires a=0, which is not physically meaningful for R. The method is validated as effective for determining the dimensional relationships among variables, affirming its reliability for future analyses.
Saladsamurai
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Homework Statement



Hi. I have a function that contains 4 variables: Q, R, \mu, dp/dx

I wish to choose 3 of them, such that they cannot be combined into a dimensionless product.I have chosen (correctly) R, \mu, dp/dx and I would like to know if my method sounds correct:

If we know the dimensions: [R]=[L] [\mu]=[ML-2T-2] and [dp/dx]=[ML-1T-1]and I know that in order for them to be dimensionless, their power product must equal zero:

[L]a[ML-1T-1]b[ML-2T-2]c=[MLT]0

or the system

a-b-2c=0
b+c=0
-b-2c=0

By inspection, this system can only be satisfied if a=0 but that does not make any sense since R is a physical quantity.

Hence I have reasoned that these 3 variable cannot form a non-dimensional parameter by themselves.Does this work? Thanks!
 
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Ignore length real quick. [MT^{ - 1} ]^a [MT^{ - 2} ]^b can NEVER be dimensionless (except for a=b=0)
 
Pengwuino said:
Ignore length real quick. [MT^{ - 1} ]^a [MT^{ - 2} ]^b can NEVER be dimensionless (except for a=b=0)

Okay cool!

Also, though longer and more tiresome, my method above works right? For the same reason.

I just want a general method just in case inspection is not that obvious.
 
Yes, that method works.
 
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